SCIAMVS 2 (2001), 47-74 The Geometrical Works of Abu Sa'Id al-DarTr. al-JurjanT

Jan P. Hogendijk

Department of Mathematics University of Utrecht

For Abu Max

I Abii Sacid al-Qarir al-JurjanI

Almost nothing is known about the life of the geometer Abu Sa<-rdal-I;)arir ("the blind") al-JurjanI. His name indicates that he originated from the town of Jurjan, that is present-day Gargan near the Caspian Sea in Iran. The most complete form of his name appears in one version of al-BirunI's Extraction of Chords [20, p.13]1 as Abu Sa<-rdMul).ammad ibn AlI al-I;)arir al-JurjanI. In [19, p. 27], Suter confused the geometer, Abu Sa<"fdal-I;)arir al-JurjanI, with the philologist, Abu Sa<-rdal-I;)arir Al).mad ibn Khalid, who died in 895 CE. 2 Consequently, al-JurjanI is listed as a ninth-century mathematician by Sezgin [17, vol. 5, pp. 263-264, vol. 6, p. 159] and by Matvievskaya and Rozenfeld [13, vol. 2, p. 76, no. 48]. In the Geometrical Problems, published below, al-JurjanI mentions Abu Abdallah al-ShannI, 3 who lived in the second half of the 10th century CE. This reference confirms that the geometer Mul).ammad ibn AlI al-JurjanI is not identical to the philologist Al).mad ibn Khalid. We will see below that al-JurjanI must have flourished around 1000 CE. 4 Two treatises by al-JurjanI have come down to us, both of considerable historical interest.

1. His Geometrical Problems begin with a construction of a neusis by means of a parabola and hyperbola. This construction is a variation of a construction in Book IV of the Mathematical Collection of Pappus of Alexandria. Then follow solutions of two problems related to the trisection of the angle, and a theorem about the altitude of a triangle. The Geometrical Problems is addressed to an anonymous mathematician, who will be identified below as al-BirunI. One of

1 0n al-BfriinI (973-1048) see for example [17, vol.6, pp.261-276]. 2 0n Abu Sa'Id al-Qarir Agmad ibn Khalid see [17, vol.8, pp.167-168]. 3 0n al-ShannI see [17, vol. 5, p. 352]. Around 970 CE he quarreled with Abii'l-Jiid in connection with the regular heptagon, see [7, pp. 243-244]. 4 In [5, p. 331] Hermelink stated that al-JurjanI lived in the second half of the 10th century CE. 48 Jan P. Hogendijk SCIAMVS 2

al-JurjanI's constructions is related to the "Problem of Alhazen" in the Optics of Ibn al-Haytham. Thus the Geometrical Problems of al-JurjanI are connected to some of the highlights of Arabic-Islamic geometry. 2. Most of al-JurjanI's Extraction of the Meridian Line is based on a lost work Analemma by Diodorus of Alexandria, a Greek expert on sundial theory who lived in the first century BCE. The only other extant trace of Diodorus' Analemma is Chapter 20 in al-BirunI's Exhaustive Treatise on Shadows.

Al-JurjanI's treatise on the meridian line was published in 1922 in a free German translation by Schoy [18], but the Arabic text has not hitherto been edited. The Geometrical Problems seem to be completely unknown to the literature. The only other traces of geometrical work by al-JurjanI are two short proofs which are cited by al-BirunI in the Extraction of Chords.5 The purpose of this paper is to publish and translate the Geometrical Problems and the Extraction of the Meridian Line, in order to make the entire extant work of al-JurjanI available for further historical analysis. I begin with a brief description of the contents of the Geometrical Problems, and a brief summary of the Extraction of the Meridian Line. This is followed by a section on manuscripts and editorial procedures, and edited Arabic texts and English translations of the two treatises.

II The Geometrical Problems of al-J urjanI

Al-JurjanI begins his Geometrical Problems with a preliminary construction which I render in modernized notation (Figure 1). The text and translation can be found below. Given: three points A, B, G on a circle, and a segment HZ. Required: to construct a straight line AED which intersects segment BG at E and the circle at D in such a way that ED = H Z. 6 In ancient Greek geometry, this type of construction was called a neusis, that is the insertion of a straight segment (DE) of given length (HZ) between two given lines which may be straight or curved (in this case segment BG and arc BG), in such a way that a given point (A) is on the segment or on its rectilinear extension. Al-JurjanI bisects BG at T and he draws a perpendicular to BG through T. He draws AL parallel to BG to meet the perpendicular at L, he drops perpendicular

5 The Arabic text is available, on the basis of the Patna manuscript, in [1, no. 1, pp. 8, 23] and [2, pp. 40, 57-48], for a facsimile of the Patna manuscript of the first proof see [2, p. 30]. A German translation of the two proofs, on the basis of the Leiden manuscript, can be found in [20, pp. 13, 15]. 6 1n this paper I use a notation such as ED in the ancient Greek way. Thus ED may indicate a straight line segment with endpoints D and E, but also the length of this segment. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-Qarir al-JurjanI 49

s

N

Figure 1

AK onto BG, and he chooses point Son KA extended such that AS= AK. 2 He then defines point N on line LT such that LN · HZ = i:BG • He assumes 2 HZ· AK< :tBG , whence LN > LT. He then draws one branch 1{ of the equilateral (orthogonal) hyperbola through K with centre A and transverse axis SK, and the parabola P with vertex N, axis NL and parameter (latus rectum) HZ. Al-Jurjan1 supposes that P and 1{ intersect at point M, and he drops perpendic­ ulars ME onto BG and M X onto AK extended. Let M X intersect LN at O. Since Mis on 1{, we have by Apollonius, Conics 1:21,7 MX 2 = SX, KX. But MX = EK, henceEA 2 = EK 2 +AK 2 = SX·KX+AK 2 = (AX 2 -AK 2 )+AK 2 = AX 2 , so EA= AX. Since Mis on the parabola P with latus rectum HZ, we have by Conics 1:118 M0 2 =ON· HZ. But MO= ET, and by definition GT 2 = LN · HZ. Thus by 2 2 subtraction BE· GE= GT - ET =LO· HZ= AX· HZ= AE ·HZ.From the geometry of the circle, BE· GE= AE · ED. Hence ED= HZ as required. Al-Jurjanl's construction can be seen as a variation on a construction in the end of Book IV of the Mathematical Collection of Pappus of Alexandria (ca. 300 CE). 9 Pappus solves the same problem by means of two conic sections P' and H', which

7 Putting XM = y, KX = x, SK= c we obtain the modern equation y 2 = x(x+c) of the hyperbola. 8 Putting OM= Y1, NO= x1, HZ= p we obtain the modern equation yf = px 1 of the parabola. 9 Pappus divides the construction into three propositions, see [23, vol. 1, pp. 231-235], [15, vol. 1, 50 Jan P. Hogendijk SCIAMVS 2

are the images of 'P and 1t under the translation of the plane which maps A to K .10 The perpendicular projection of the intersection M' of 'P' and 1t' on line BG is of course point E in Figure 1. The similarity with Pappus suggests that al-JurjanI's construction is ultimately of ancient Greek origin. We now turn to the diorismos of the problem, that is to say, the necessary and sufficient condition for the existence of a solution. 11 Al-JurjanI wrongly states that 2 the diorismos is HZ · AK :S f BG . Actually, this condition is necessary but not 2 sufficient. If HZ· AK = f BG , the problem can only be solved if K is the midpoint of BG. Figure 2 is drawn for the case where K is not the midpoint of BG and 2 HZ· AK is slightly less than f BG , so that N is a little below T. In this case 1t and 'P do not intersect, and the problem has no solution.

L >----...... _A

H Z

p

Figure 2

If K is not the midpoint of BG, the correct diorismos has the form HZ :S C for some segment C with C · AK < f BG 2 . If HZ = C, the curves 'P and 1t are tangent, and by analyzing this case one can in principle find a construction of C from AK, BK and BG. This problem is by no means easy: the construction of line AED itself is equivalent to the solution of an irreducible algebraic equation of degree 4, and the length of C is the root of an irreducible cubic equation. By means of a different construction, lbn al-Haytham and al-Mu'taman ibn Hud (died 1085) found C as a normal from a point to a certain hyperbola. 12

pp. 298-303, corrected in vol. 3, pp. 1231-33]. 10 1 use the concept of "translation" for sake of brevity. Of course I do not want to suggest that Pappus or al-Jurjan1 had the modern concept of a translation of the plane. Note that point K is the centre of rt', and that P' passes through points B and T because BT 2 = LN · HZ. 11 Pappus does not concern himself with any diorismos to this problem. On the concept of diorismos in general see [4, vol. 1, pp. 130-131], ancient examples are e.g. Elements 1:22 [4, vol. 1, p. 239] and the ancient additions to Archimedes, Sphere and Cylinder 11:4, see [3, p. 197]. 12 Ibn al-Haytham considered the case where BG is a diameter [16, p. 318], and al-Mu'taman found the generalisation to arbitrary positions of BG [8, p. 77]. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-I;)arir al-JurjanI 51

Al-JurjanI says that the person to whom he addresses these problems had said, on the authority of al-ShannI, that the problem does not require the condition HZ· AK :S :tBG 2 . Al-ShannI may have thought about solutio~s ADE for which point E is on BG extended outside the circle. These solutions can be found for any choice of the segment HZ, by means of the intersections of P with the other branch of the hyperbola 1{. In the second problem, al-JurjanI considers points B and G on the arms of a given angle BAG such that AB= AG (Figure 3). Required: to construct point Z 2 2 on segment AB such that GZ ·AB+ AZ = AB . The solution is easy: Al-JurjanI draws the circle with centre A and radius AB, he extends BA to meet the circle again at point D, and he constructs, by the method of his first problem, the straight line GZE meeting BA at point Z and the circle at point E such that EZ = AB. 2 2 2 2 Then GZ ·AB+ AZ = GZ · EZ + AZ = BZ ·DZ+ AZ = AB , as required. Al-JurjanI uses the theorems GZ · EZ = BZ · DZ and AB 2 -AZ 2 = BZ · DZ from Euclid's Elements.

(P)

E \ (Q) ' ' " " •: (R)

Figure 3

This problem makes it possible to identify the mathematician to whom al-JurjanI sent this treatise. In a work on the trisection of the angle by the late tenth-century geometer al-SijzI, the problem is mentioned as a "lemma by al-BirilnI'' for the tri­ section of the angle [24, p. 119].13 Al-BirilnI posed this problem to Abil'l-Jud, who then gave a solution by means of a parabola and a hyperbola which does not in-

13 To see the connection, consider in Figure 3 P AQR perpendicular to BAD, and extend CZ to

0 meet PQ at R. By al-JurjanI's construction, GA = AE = ZE, and because LEAR = go , AE = ZE = ER. Put LARZ = x. Then LRAE = x, hence LACZ= LAEZ = 2x, so LGAP = 3x.

0 Thus LGAP = go - LGAB can be trisected by means of this lemma. 52 Jan P. Hogendijk SCIAMVS 2 volve al-JurjanI's first problem [24, pp.114-115]. Thus the mathematician to whom al-JurjanI sent his Geometrical Problems was probably al-BirunI. Abu'l-Jud flour-. ished around 970 CE, so he was at least one generation older than al-BirunI, who was born in 973 CE, and therefore their correspondence probably took place by the end of the 10th century or around 1000 CE. It is likely that al-BirunI corresponded at the same time with various people about similar problems, so a date around 1000 CE seems to be most probable for the Geometrical Problems of al-JurjanI as well. In his third problem (Figure 4), al-JurjanI considers a given triangle ABG with right angle A, and a given point D on the hypotenuse. Required: to construct point Z on AD in such a way that if G Z E is drawn and extended to meet AB at point E, GE: AZ = GB: AD. Al-JurjanI does not give any motivation for this odd problem. If AD: BG = 1: 2, the solution of the problem can be used for the trisection of an angle. See Figure 4, where BG= 2AD, EG = 2AZ, C is the midpoint of EG, and !_EGA= } LBG P .14 In his work on the trisection of the angle, al-SijzI mentions this problem, with AD: BG = 1: 2, as one of five problems to which al-BirunI had related the trisection of the angle [24, p.124]. This confirms my conclusion that al-JurjanI sent his Geometrical Problems to al-BirunI.

E

G A Figure 4

I now turn to the connection between al-JurjanI's solution to the third problem, and the "Problem of Alhazen" in the Optics of Ibn al-Haytham. First it is necessary to summarize al-JurjanI's proof. Assume that the ratio AD : BG is arbitrary (Figure 5). Al-JurjanI constructs the circumscribed circle of triangle ABG and he extends AD to meet this circle at point T. Using his first construction, he draws the straight line T LK to intersect line BG at point L and the circle at K such that LK = AD. He draws KB and KG. From the geometry of the circle he deduces LBK L =

14 Since LEAG in Figure 4 is a right angle, EC = CG = CA, so triangle CZA is isosceles. If we put LBGA = a, LEGA = (3, we can express the three angles of triangle EZA in terms of a and (3, and thus arrive at the conclusion a + 3(3 = 180°. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-Qarir al-JurjanI 53

LBKT = LBAT = LBAD and similarly LGKL = LGAD. Al-JurjanI now draws line MDS such that LMDA = LBLK. Since KL= AD, the figures BKLG and MADS are congruent, so MS= BG. He then draws GZE para,llel to SDM. By similar triangles EG: AZ= MS: AD so line GZE is the required solution.

Figure 5

In the course of his proof, al-JurjanI notes that he has now solved the following problem: to construct through a given point D a straight line (M DS) which in­ tersects two given lines (AB and AG) at points M and S in such a way that the segment MS has a given length (equal to BG). This problem is a neusis between two straight lines (AB and AG). Al-JurjanI's first problem was a neusis between an arc and a chord of a circle. 15 What al-JurjEnI has shown is how the neusis between the two straight lines can be reduced to the neusis between the chord and the arc of a circle. The "problem of Alhazen" is a problem about spherical, cylindrical and conical mirrors, which was solved by al-I:{asan ibn al-Haytham (ca. 965-1041) 16 in Book V of his Optics. The Optics were translated into Latin in the twelfth century and thus the problem became famous in the West. In this problem one assumes that the positions of the eye and the image are given, and one is required to construct the points of reflection in the mirror. Ibn al-Haytham knew that there is a maximum of four points of reflection. The problem is equivalent to an algebraic equation of degree four. Ibn al-Haytham reduces the "problem of Alhazen," through a series of compli­ cated preliminaries, to two neusis-constructions involving conic sections, namely Optics V:33 (Lemma 1) and Optics V:34 (Lemma 2), which have been translated by Sabra [16, pp. 303-309, 315-320]. In Lemmas 1 and 2, he considers a given point A on a circle with given diameter BG, and he constructs line AED which intersects

15 Note that the given point should be on the same circle. 16 0n Ibn al-Haytham see e.g. [17, vol.5, pp.251-261; vol.6, pp.358-374]. 54 Jan P. Hogendijk SCIAMVS 2 the circle in D and BG at point E, such that DE is equal to a given segment HZ, and such that E is outside the circle in Lemma 1, and inside the circle in Lemma 2 (for which see Figure 6). I have pointed out previously that his construction of Lemma 1 is similar to a neusis-construction by means of a hyperbola and a circle, which is related to the trisection of an angle, and which is found in Book IV of the Collection of Pappus of Alexandria and in treatises by the Banu Musa ( ca. 850) and Thabit ibn Qurra (836-901) [8, p. 62]. Ibn al-Haytham's solution of Lemma 2 is essentially different from Lemma 1, and this is odd because the problems are so similar. 17 However, I will now show that Lemma 2 is closely related to al-JurjanI's solution of his third geometrical problem. Al-JurjanI shows how a neusis between two given straight lines can be reduced to a neusis between a chord and an arc of a circle. lbn al-Haytham uses this relationship, but he argues the other way around. Thus he shows how the neusis between the chord and arc of a circle can be reduced to the neusis between the two straight lines. To illustrate the relationship, I have redrawn al-JurjanI's figure as Figure 6, with notations used by lbn al-Haytham in the translation by Sabra [16, pp. 318-320]. The situation is complicated by the fact that lbn al-Haytham uses two figures. The letters A, B, G, D, E occur in his original figure, and [Z], [HJ, [L] and [M] in his auxiliary figure. The auxiliary figure is congruent with part of the original figure and I have added the square brackets to avoid confusion. 18

D'

[L']

Figure 6

lbn al-Haytham argues as follows (Figure 6). Suppose that point A is a given

17 1n his simplified solution of the "problem of Alhazen," the AndalusI mathematician al-Mu'taman ibn Hud solved lbn al-Haytham's Lemma 2 by the same method as Lemma 1, cf. [8]. 18 The reader who is still confused by the original and auxiliary figure should redraw the auxiliary figure on a separate sheet. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-Qarir al-JurjanI 55 point on a circle with given diameter BG. Let the segment HZ be given. Required: to construct a straight line AED which intersects the diameter at E and the circle at Din such a way that ED= HZ. lbn al-Haytham notes that, although the position of AED is not given, we have by an elementary theorem on the geometry of the circle LADG = LABG and LADE= LAGE, so LADG and LADE are known. He then draws in an auxiliary figure the segment HZ and two lines H L and HM such that LLHZ = LABG and LMHZ = LAGE (hence LLHM = 90°). Note that the precise positions of point L on line H L and point M on line HM are still to be determined. Ibn al-Haytham now constructs in his auxiliary figure a straight line MZL, with point M on line HM and L on line H L, in such a way that LM = BG, where BG is a segment in the original figure. In this construction, which does not concern us here, he uses the intersection of another circle and a hyperbola. Thus he has now solved the neusis between two straight lines. In his original figure, he now constructs point D on the circle such that LG B D = LLMH and he draws DEA and DC. Then figure LZMH in the auxiliary figure is congruent to figure GEBD in the original figure, whence ED= ZH as required. Thus he has now reduced the neusis between the arc and chord BG to the neusis between the straight lines M H and H L. Ibn al-Haytham renders only one solution in his figure but he is well aware of the fact that there may be a second solution. I have indicated the second solution by E', D', M', L'. The reasonings of lbn al-Haytham and al-JurjanI differ to the extent that al-JurjanI wants to construct a second solution of the neusis when one solution is already given. This is why lbn al-Haytham needs an auxiliary figure, in which his given segment HZ corresponds to the given line AD in al-JurjanI's Figure 5. For the rest, the reasonings by al-JurjanI and lbn al-Haytham are so closely related that they can hardly be independent, and thus the question arises who influenced whom. We have seen that al-JurjanI's diorismos for the neusis between the chord and arc of a circle is incorrect, but lbn al-Haytham was able to find the correct diorismos by means of the circle and hyperbola in Lemma 2 in his construction of the neusis between two given straight lines. If al-JurjanI had seen this construction, he would not have given a wrong diorismos for the neusis between the chord and arc of a circle. I conclude that al-JurjanI influenced lbn abHaytham and not the other way around. lbn al-Haytham's solution of the "Problem of Alhazen" is one of the high points of mathematics in Islamic civilization, and it now turns out that al-JurjanI was an important source of inspiration. The fourth proposition in al-JurjanI's Geometrical Problems concerns an arbitrary triangle ABC with its altitude AD. Al-JurjanI proves an identity which can be used to determine BD in terms of the sides AB, AG, BG. The proof is elementary. 56 Jan P. Hogendijk SCIAMVS 2 III The Extraction of the Meridian Line

The Book of the Extraction of the Meridian Line Prom the Book Analrma, and the Proof For It by Abu SaTd al-J)arfr was first published in 1922 in a German translation by Schoy [18]. I have divided the text into four sections, numbered (1), (2), (3) and (4). Sections (1)-(3) deal with the construction of the north-south line on a horizontal plane, by means of three unequal shadows cast by a vertical stick at three different moments during the same day. Section (4) presents a method for finding the cardinal directions by means of one shadow measurement at the right moment. In the northern hemisphere, the method can only be used during the spring and the summer. Section (4) of the text seems to be an addition by al-JurjanI, whereas Sections (1) to (3) appear to be based on the Book Analfmii, mentioned in the title. Hl Although Schoy could not identify this book, he pointed out that the method in Sections (1 )-(3) is of ancient origin and that it was also used by the Roman surveyor Hyginus (ca. A.D. 100). In 1959, Kennedy published a construction of the meridian line by means of three shadows in Chapter 20 of the Exhaustive Treatise on Shadows by al-BirilnI [9]. Al-BirilnI says that the method was taken from the "Book Analima" of Diodorus of Alexandria, a Greek mathematician from the first century BCE. Heinrich Hermelink [6] then pointed out that the methods in [18] and [9] were the same, so the "book Analima" in the title of al-JurjanI's treatise is the Analemma of Diodorus, now lost. Thus al-BirilnI and al-JurjanI preserved two versions of a construction by Diodorus. The two versions have been further explained and compared by Neugebauer [14, vol. 2, pp. 840-843] and Kennedy [10, vol. 2, pp. 91- 93], and I see no need to repeat what has been said in these publications. I will only compare the structure of Sections (1)-(3) of al-JurjanI's text, in his notations, with Chapter 20 of the Exhaustive Treatise on Shadows by al-BirilnL The Arabic text of this chapter is available in [1, no. 2, pp.116-119, 121], for an English translation see [10, vol. 1, pp.162-166]. Al-JurjanI considers a gnomon AB with tip B and foot A, which casts three unequal shadows AG > AD > AE in the horizontal plane (see Figure 11 below). He defines points Hand Ton lines BD and BG such that BH =BT= BE, and he drops perpendiculars H L and T K onto the horizontal plane. He then defines S as the point of intersection of KL and GD. In (1) he proves that ES is parallel to the East-West line. He then shows in (2) that GD and KL are never parallel, so the intersection S always exists. In (3) he presents the practical construction of S from the positions of G, D and E and the lengths of BG, BD and BE. Al-BirilnI presents these elements in a different order. He begins with the practical

19 1n [13), the treatise by al-JurjanI is incorrectly listed as a commentary on the Analemma of Ptolemy. SCIAMVS 2 The Geometrical Works of Abu Sa'1d al-Qarfr al-JurjanI 57 construction (3) and he also explains how the lengths of BG, BD and BE are to be found from those of AB, AG, AD and AE. In the proof of the correctness of the construction, al-BirunI presents the argument (2) in the same way as al-JurjanI but before the definition of point S. Al-BirunI uses a different notation, but there are some striking linguistic simi­ larities between his explanations and those of al-JurjanI. So it is clear that the two authors were excerpting the same work. The gist of (4) is as follows in modern notation. For solar declination <5 > 0, geographical latitude

IV Manuscripts and Editorial Procedures

The two extant texts by al-JurjanI have come down to us in a single Arabic manuscript, namely Cairo, Dar al-Kutub, Mu~tara Fa<;lil Riya<;la 41m. The Geo­ metrical Problems are in ff. 69b-71a and the Extraction of the Meridian Line is in ff.153b-154b, see [11, p. 36, no. B24] and [12, vol. I, pp. 443, 445; vol. 2, pp. 833- 834]. The codex Mui;;tafa Fa<;lilRiya<;la 41m was copied around 1153 H./1740 CE by Mui;,tara $idqI ibn $alil}.. In [13, vol. 2, pp. 76-77], Rosenfeld and Matvievskaya men­ tion a Berlin manuscript of the two treatises by al-JurjanI. This "Berlin manuscript" was a modern copy, now lost, of the Cairo manuscript. For easy reference I have divided each of the two treatises into four sections. In my edition of the Arabic texts I have added the interpunction, and I have made some changes in orthography without noticing this in the apparatus. Thus I have changed ~, li1=,and~ to ~Lo,~ and L:..lA.In his translation of the second text [18], Schoy made a number of corrections to make mathematical sense. In my apparatus to the Arabic text I have indicated these corrections by the notation Sch. There are a few passages which Mui;,tafa $idqI did not understand, but on the whole the quality of his manuscript is very good. I have redrawn the figures but have tried to stay close to the figures in the manuscript. My own additions to the text or translation are in angular brackets < > or in parentheses ( ) . 58 Jan P. Hogendijk SCIAMVS 2 V The Geometrical Problems: Edition

~ Jw .JJI "i..P_JJ~_}.l- ..1-~I ~ 0-~ ~~ yL....i

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39 40 41 42 ~ ~.): _f"..>dl> MS. ~ ~.): _f"..>dl> MS. ~Sch.: ~ MS. The MS. repeats ~..,~ J!-4... .!It.. ~- . SCIAMVS 2 The Geometrical Works of Abu Sa'1d al-Qarir al-JurjanI 63 VII The Geometrical Problems: Translation

In the name of God, the merciful, the compassionate.

Geometrical Problems by Abu Sa<"fdal-I;arfr al-JurjanI, may God Most High have mercy upon him.

< 1 > (Figure 7) Let there be a circle ABDG. Chord BG is known. 43 We want to draw a line such as line AE D in such a way that line ED is equal to an assumed line, and let the assumed line be line Z H. We extend line BG on both sides, and we drop from point A perpendicular AK to the line. If the product AK times HZ is greater than one quarter of the square of BG, the problem is impossible, but if it is equal to or less than one quarter of it, it is possible. That is, the product AE times ED is not greater than a quarter of the square of BG since it is equal to the product BE times EG, and the product AE times ED is greater than (or equal to) the product AK times ED, which is equal to line ZH. 44 I mention this because you said on the authority of Abu 'Abdallah al-ShannI, may God support him, that this condition is not necessary. Now I return to the presentation of the rest of the problem. I say: We bisect line BG at point T. We erect at point Ta line perpendicular to line BG and we extend it on both sides, and it is line NTL. We draw from point A line AL parallel to line BG. We cut off from line LT N line LN and let LN times ZH be equal to the square of BT. 45 Then it is clear that point N is located under line BG since the product AK times ZH is less than the square of BT and TL is equal to AK, so LN is greater than LT. We extend AK on the side of A towards S and we make AS equal to AK. We extend line SAK indefinitely on the side of K. We construct a hyperbola with vertex point K, transverse axis SK, and latus rectum also equal to SK. 46 We construct a parabola with vertex point N, diameter line LTN and latus rectum line ZH. 47 Then these two (conic) sections intersect at point M. 48 We draw from M line MOX parallel to line BG and also from point M line MQE parallel to line SK. Then, as a consequence of the twenty-first proposition of the first Book of the Conics,49 the product SX times XK is equal to the square of MX. We add the square of KA, then the product SX times XK plus the square of AK is equal to the

43 Al-JurjanI also assumes that point A is known. 44 2 This argument shows that the condition AK · ZH :<:::; iBG is a necessary condition for the existence of a solution. The condition is not sufficient, see the commentary above. 45 For the construction of point N see Euclid, Elements I:44-45 [4, vol. 1, pp. 341-347]. 46 See Apollonius, Conics I:54 [22, p.101]. 47 See Apollonius, Conics I:52 [22, p. 97]. Line LT N is the axis of the parabola. 48 This is not necessarily true, see the commentary above. 49 For Conics I:21 see [22, p. 43]. 64 Jan P. Hogendijk SCIAMVS 2 square of MX plus the square of AK. But MX is equal to EK. Thus the square of AE is equal to the square of AK and (i.e. plus) the product SX times XK, and (therefore) 50 the square of AE is equal to the square of AX. So AX is equal to AE. But since (conic) section MN is a parabola with diameter NL and latus rectum line ZH, the product ZH times ON is equal to the square of ET, 51 which is equal to MO. The product ZH times LN is equal to the square of BT, but the product ZH times NO was shown to be equal to the square of ET. Therefore the product ZH times LO, which is the remainder, is equal to BE times EG, 52 since line BG is divided into halves at T and into unequal (parts) at E, so the product BE times EG together with the square of ET is equal to the square of TG, 53 which is equal to TB. But the product ZH times NO is equal to the square of ET, which is equal to the square of OM, and LN times HZ was equal to the square of TG. So if we subtract from the product ZH times LN the product ZH times ON, and from the product BE times EG plus the square of ET the square of ET, the remainder will be the product LO times ZH, equal to the product BE times EG. But LO is equal to AE since it is also equal to AX, so the product AE times ZH is equal to the product BE times EG. But the product BE times EG is equal to the product AE times ED, 54 so ED is equal to the assumed (line) ZH. That is what we wanted to prove. s

N Figure 7

50 AK 2 + SX · XK = AX 2 by Euclid, Elements 11:5, see [4, vol. I, p. 382]. 51 Apollonius, Conics 1:11 [22, p. 21] 52 The following sentences of this paragraph explain or repeat what has already been said. 53 Euclid, Elements 11:5 [4, vol. 1, p. 382]. 54 Euclid, Elements 111:35 [4, vol. 2, p. 71J. SCIAMVS 2 The Geometrical Works of Abu Sa'"id al-Qar1r al-Jurjan1 65

< 2 > (Figure 8) Lines GA, AB are assumed and contain an angle BAG. We want to draw from point Ga line, say GZ, such that the product GZ times AB plus the square of AZ is equal to the square of AB. 55 Let us extend AB towards D and we make AD equal to AB. We join DG and we construct about triangle BDG circle BEDG. Then circle BEDG is known in position, point G on it is known, and segment BED of this circle is assumed. Then we join from point G a straight line towards arc BED such that the straight line which falls between this arc and its chord BD is equal to AB. Thus let it be drawn, and let that (line) be GZE. I say that GZ times AB plus the square of AZ is equal to the square of AB. Proof of this: GZ times ZE is equal to the product BZ times ZD. 56 We add the square of ZA, then the product GZ times ZE plus the square of ZA is equal to the product BZ times ZD plus the square of ZA, that is to say, equal to the square of AB. Thus the product G Z times Z E plus the square of Z A is equal to the square of AB. But ZE is equal to AB. Thus the product GZ times AB plus the square of AZ is equal to the square of AB, and that is what we wanted to prove.

E Figure 8 < 3 > (Figure 9) Triangle ABG is right-angled and its angle A is the right angle, and it 57 is known in magnitude and in shape. A line AD has been drawn in it and the ratio of BG to AD is a known ratio. How do we draw from point Ga line GZE such that the ratio of GE to AZ is equal to the ratio of BG to AD? 58 This problem requires that lemma which we have mentioned, because it requires that we draw from point D a line such that the part of it which falls between lines AB and AG is equal to line BG. By means of the lemma which we have mentioned, this and other things can be drawn, in a right-angled triangle as well as in a triangle

55 Al-Jurjan1 assumes GA = AB. 56 Euclid, Elements 111:35 [4, vol. 2, p. 71]. 57 According to the manuscript, the angle is known in magnitude and in shape. I have emended the Arabic hiya to huwa. The emended text means that the triangle is known in magnitude and in shape. 58 The scribe wrote ~rand A6D in the margin of the manuscript. In his figure, BG : AD is approx­ imately 14: 6. 66 Jan P. Hogendijk SCIAMVS 2 which is not right-angled. The reason is that the circumscribed circle of triangle ABC is a known circle. Thus let us circumscribe about this triangle circle BTG A. Let us extend line AD towards point T, then point Ton the circle is known, and segment BAG is a known segment. So let us draw from point T a straight line towards arc BAG in such a way that the part of it which falls between arc BAG and chord BG is equal to AD. If such a line can be drawn other than TD, the problem is solvable, but if it cannot be drawn, it (the problem) is impossible, and this is easy (to verify) for someone who thinks (about it) and who uses in it the lemma which we have mentioned. 59 Thus let us draw this line TLK, and let LK be equal to AD. We join BK, GK, then it is clear that angle BKT is equal to angle BAD because they stand on the same segment. 60 Thus we construct on line AD a triangle similar to triangle BK L; it is clear that one of its sides is part of line BA. Let it be triangle AM D. Let us extend line MD to meet line AG at point S. Then I say that we have drawn through point D a straight line MS such that the part of it which falls between lines AB, AG is equal to line BG. Proof: Triangle DAM is similar to triangle BK L and line KL is equal to line AD, so line MD is equal to BL. In the same way we also prove that DS is equal to GL. Thus MS is equal to BG, and we have drawn from point D a line MDS equal to BG. So let us now draw from point G line G Z E parallel to MS. I say that G Z E is what was desired. Proof: We draw from point Z line FZQ parallel to line BG. Then the ratio of BG to AD is equal to the ratio of FQ to AZ, and the ratio of MS to AD is equal to the ratio of GE to AZ, but MS is equal to BG, so the ratio of BG to AD is equal to the ratio of GE to AZ, and that is what we wanted to prove.

Figure 9

59 Following a suggestion of Professor 0. Naises, I have emended ta'ammala in the manuscript to yata'ammalu. Without emendation, the text means: " ... who thought (about it). In this (construction) the lemma which we have mentioned is being used." 60 Euclid, Elements 111:27 [4, vol. 2, p. 58]. SCIAMVS 2 The Geometrical Works of Abu Sa'1d al-Qarfr al-Jurjan1 67

< 4 > (Figure 10) If in triangle ABG its altitude AD is drawn and if BA is extended in a straight line towards E such that AE is equal to AG: the excess of twice the rectangle EB times BA over twice the rectangle GB times BD is equal to the excess of the square of BE over the square of BG. Proof of this: Twice the rectangle BE times BA is equal to twice the square of BA and twice the rectangle BA times AE. Twice the rectangle BG times BD is twice the square of BD and twice the rectangle BD times DG. 61 But the excess of twice the square of AB over twice the square of BD is twice the square of AD. Thus the excess of twice the square of AB plus twice the rectangle BA times AE over twice the square of BD plus twice the rectangle BD times DG is twice the square of AD plus the excess of twice the rectangle BA times AE over twice the rectangle BD times DG. But the square of BE is equal to the (sum of the) squares BA, AE plus twice the rectangle BA times AE, and the square of BG is equal to the (sum of the) squares BD, DG plus twice the rectangle BD times DC. The excess of the square of AB over the square of B D is equal to the square of AD, and the excess of the square of AE, which is equal to AG, over the square of GD is equal to the square of AD. Thus the excess of the square of BE over the square of BG is equal to twice the square of AD and twice the excess of the rectangle BA times AE over the rectangle BD times l)G, and this is equal to the excess which we have obtained before. That is what we wanted to prove. End of the treatise.

E E

A

G D B D G B

Figure 10

On Monday the 29th of Jumada I of the year 1153.62

61 The manuscript has two figures, one for the case where D is between B and C (as in the text), 2 and one for the case where C is between Band D. In that case 2BC ·ED= 2BD - 2BD · DC 2 2 2 2 2 and BC = BD + DC - 2BD · DC, and also BE - BC = 2BE · BA- 2CB · ED. I leave the details to the reader. 62 The date of copying the manuscript corresponds to August 22, 1740 CE. 68 Jan P. Hogendijk SCIAMVS 2 VIII The Extraction of the Meridian Line: Translation

In the name of God, the merciful, the compassionate.

The book of the extraction of the meridian line from the book Analima, and the proof for it by Abu Sa

< 1 > (Figure 11) Let three successive shadows of the gnomon be observed, let the longest of these be AG, the intermediate one AD, and the shortest one AE, and let the gnomon be AB. We join BG, BD, BE and we describe with centre Band radius BE arc EHT. 64 We drop perpendicular TK onto AG and perpendicular HL onto AD. 65 We join KL and GD and we extend them to meet at point S. 66 Then we join SE, it is parallel to the East-West line, and the perpendicular drawn to it from point A is part of the meridian line. Proof of this: It is clear that the intersection of the plane of circle EHT and the plane of the line which is described by the shadow in the horizon plane is parallel to the equator, 67 and (it is clear) that line HT is the intersection of three planes: the first is the plane of circle EHT, the second the plane HTL and the third the

63 The simplified procedure is part 4 of the text. 64 Points Hand Tare supposed to be on ED and BG, so the circle is the base of a cone with vertex B. The cone is described by the rays through the centre of the sun and the tip of the gnomon in the course of one day. The plane of circle EHT is parallel to the celestial equator. 65 Here I have followed Schoy's emendation. Mu~tafa $idqI wrote: "We drop perpendicular T K onto BG and perpendicular HL onto BD." In his figure line BG is perpendicular to TK, line BD is perpendicular to H L, and points K, L are not on lines AG, AD respectively. Schoy's emendation is consistent with the proof in (2) that KL intersects GD. Mu~tafa $idqI made so many scribal errors in this proof (compare the critical apparatus) that Schoy was unable to make sense of the passage, see [18, p. 266, fn. l]. 66 Note that line KL is only used for the definition of point S and in the proof that point S exists. Line KL does not play a role in the practical construction of S in section 3 of the text. 67 The text is confused. It would be correct to say that the plane of circle EHT is parallel to the plane of the celestial equator, so its intersection with the horizontal plane is parallel to the East­ West line (Arabic: khatt al-i'tidiiQ. Compare the corresponding confused passage in al-BirilnI's summary of Diodorus' Analemma [l, no. 2, pp. 118:13-119:3]: ~~- ':?lll ~I ~ ~-' JJ,JI .1.,_,)'tr ~ 4ru o)b .f ~ ~ ~..,u..11J_, .•.uJI 01 i _f...o_, .JlrJI '-4..Ai ..b ~ ~I tr-' ..i ~JI_,..o)l~I 0'J .JlrJI J..\a.o~ ~JI_,..0§.; ._;;'JI _) JJ,JI v-L "It is known that the intersections of the plane of every circle perpendicular to the axis of the shadow cone with the plane of the (conic) section which the tip of the shadow produces in the horizon are parallel to the plane of the equator since the circle is parallel to it, and ( that ?) the axis SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-Qarir al-JurjanI 69 plane HTG D. Line KL is in plane HT KL, but line GD is in plane HTG D, and the meeting (point) of the two (lines) is at point S. Thus point S is at the intersection of the three planes we mentioned. So it is clear that line TH S is a straight line, since it is the intersection of the above-mentioned planes. Again, point S is at the intersection of the two planes HT KL, HTG D, and this intersection is also the intersection of the two above-mentioned planes and the plane of circle EHT. 68 Thus point S is in the plane of this circle and in the plane of the horizon, so it is at the intersection of the plane of the horizon and the plane of this circle. But point E is also in the plane of the horizon and in the plane of circle EHT, and we have proved that point Sis also in these two planes. So line SE is the intersection of the plane of the horizon and the (plane of the) circle we have mentioned, and it is parallel to the East-West line.

B

G

s

Figure 11

< 2 > Proof that lines GD, LK meet on the side of points L, D: Since line BG is longer than line BD and line BT is equal to BH, TG is longer than DH. So of the (conic) section is the meridian line" (cf. the translation in [10, vol. I, p.164]). It seems to me that al-JurjanI and al-BirunI tried to make sense of a complicated and perhaps confused passage in the Analemma of Diodorus. 68 Al-BirunI's proof ([10, p. 166], [1, no. 2, pp. 117-119]) corresponds to this proof by al-JurjanI minus the sentence "Again ... circle EHT." This sentence is mathematically redundant and probably did not belong to Diodorus' Analemma. 70 Jan P. Hogendijk SCIAMVS 2 the ratio of GT to TB is greater than the ratio of DH to HB. Componendo, 69 the ratio of GB to BT is greater than the ratio of DB to BH. But the ratio of GB to BT is as the ratio of GA to AK, and the ratio of DB to B H is as the ratio of DA to AL. Thus the ratio of GA to AK is greater than the ratio of DA to AL. So if we draw from point Kline KO parallel to line GD, point O falls between points A and L. Thus angles 0KG, DGK are (together) equal to two right angles, but they are (together) greater than (the sum of) angles DGK, GKL, so (the sum of) angles DGK, GKL is less than two right angles. Therefore lines GD and LK will meet 70 in the direction we have indicated, and that is what we wanted to prove. < 3 > (Figure 12) Again, we redraw the three shadows in the same way, and we describe in the plane of the horizon with centre G and radius GB arc ZM, and in the same way, with centre D and radius DB arc Z0. 71 Then they meet at Z. We join GZ, DZ and we cut off from lines ZG, ZD (two segments) ZT, ZH equal to line BE. We join TH, GD and we extend them to meet at S. I say that this point corresponds exactly to point S in the first figure. Proof of this: 72 Line GZ is equal to GB in the first (figure), and DZ in this (figure) is equal to DB in that (figure), and each of the lines Z H, ZT here is equal to each of.the lines B H, BT in that (figure). Thus the sides of triangle G Z D here are equal to the corresponding sides of triangle GD B there. 73 So angle G Z D is equal to angle GED there, lines TZ, ZH are equal to lines TB, BH respectively, so HT here is equal to HT there. But TG here is equal to TG there, and similarly HD is the same in both (figures), and angles ZGD, ZDG here are equal to angles BGD, BDG there. Therefore HT here meets GD at S and line GS here is equal to line GS there, since GT in both (figures) is equal, and angle ZGD here is equal to angle BG D there, and angle GT H here is equal to angle GT H there, so line TH meets GD at a point whose distance to G is equal to the distance to G of the corresponding point in the first figure, and that is point S. That is what we wanted to prove.

69 See Euclid, Elements V:18 [4, vol. 2, p. 169]. 70 Lines GD and KL meet by Euclid's parallel postulate, see [4, vol. 1, pp.155, 202-220]. 71 Presumably, the shadows AG, AD, AE are measured in the horizontal plane, and BE, BD, BG are then constructed from the length of the gnomon AB by the theorem of Pythagoras. For sake of clarity, I have drawn as dashed lines all lines in the figure which are not mentioned in the text. 72 The proof could have been shortened. It follows from the construction of the figures that triangles BCD and ZGD are congruent, and also triangles ETH and ZTH. Thus GT and angles TGD and TGH are the same in both figures, so the triangles GTS in both figures are congruent. Perhaps the proof was revised by a scribe. 73 "Here" and "there" are used in a similar way by al-ffirun1 in his paraphrase of Diodorus' con­ struction, cf. [1, no. I, pp.119:9-12]. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-I;arir al-JurjanI 71

0

B r<-, 11 ' .._..., •' '' ...... I

' ' ' ' T

'

' ' H ' G

I

A \c= = ~:: ======- 'I I I D \ I \ I \ I \ I \ I \ I \ I \I \\ "----- E s

Figure 12

< 4 > (Figure 13) For the extraction of the meridian line, let circle ABGD be on a plane tablet, (the circle being) divided into quadrants by means of diameters AEG, BED, and its circumference (being) divided into three hundred and sixty degrees. We take arc AZ in the amount of the latitude of the locality, we join EZ, and we drop perpendicular AH onto E Z. We assume AT equal to the declination of the sun on that day, and we drop perpendicular T K onto AE. We cut off from AH (segment) AL equal to KT, and we draw through L perpendicular LM towards AH, to meet AE at point M. We construct on line AE semicircle AN E. We draw AN equal to AM and we join NE. Then we cut off from line AN line AS with the magnitude of the gnomon, and we draw perpendicular SO to meet line AE at 0. Then we describe in the plane of the horizon a circle at the place we want, with magnitude (defined by) the radius SO. We place at its centre as a gnomon a perpendicular to the plane of the horizon, with magnitude AS. Then we observe its shadow before noon or after noon, until it falls on the circumference of this circle. We join its location with the foot of the gnomon by a straight line: it is the East­ West line. We draw from the centre a perpendicular to that line; then it is the 72 Jan P. Hogendijk SCIAMVS 2 meridian, and that is what we wanted to construct. God Most High knows best. The end.

A

E

G

Figure 13

On Saturday the 21st of the month Rahr II of the year 1153.74

74 The date of copying the manuscript corresponds to July 16, 1740 CE. SCIAMVS 2 The Geometrical Works of Abu Sa'Id al-Qarir al-JurjanI 73 Acknowledgement

It is a pleasure to thank Professor David King (Frankfurt) for making the Arabic manuscripts of the two treatises by al-JurjanI available to me, and for his sugges­ tion (in [11, p. 36]) to publish these treatises. I am grateful to Dr. Richard Lorch (Miinchen) for his assistance with ArabTeX, and to the anonymous referees and Prof. Dr. 0. Naises (Alexandria) for their helpful comments on an earlier version of this paper.

References

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[12] David A. King, A Catalogue of the Scientific Manuscripts in the Egyptian Na­ tional Library. Cairo: General Egyptian Book Organization, 1981-1986, 2 vols. [in Arabic] [13] Galina P. Matvievskaya, Boris A. Rozenfeld, Matematiki i astronomy musul­ manskogo srednevekovya i ikh trudy (VIII-XVIII vv.). Moscow: Nauka, 1983, 3 vols. [14] Otto Neugebauer, A History of Ancient Mathematical Astronomy. New York: Springer, 1975, 3 vols. [15] Pappi Alexandrini Collectiones quae supersunt, ed. Fridericus Hultsch. Reprint ed., Amsterdam: Hakkert, 1965. [16] Abdelhamid I. Sabra, lbn al-Haytham's Lemmas for Solving "Alhazen's prob­ lem", Archive for History of Exact Sciences 26 (1982), 299-324, reprinted in: A. I. Sabra, Optics, Astronomy and Logic: Studies in Arabic Science and Phi­ losophy. Aldershot: Variorum, 1994. [17] Fuat Sezgin, Geschichte des arabischen Schrifttums. Vol. 5: Mathematik bis ca. 430 H., Vol. 6: Astronomie bis ca. 430 H., Vol. 8: Lexikographie bis ca. 430 H. Leiden: Brill, 1974-1982. [18] Carl Schoy, Abhandlung iiber die Ziehung der Mittagslinie dem Buche iiber

(Received: August 6, 2000)